If a,b,c are in A.P. ; b,c,d are in G.P & $\frac{1}{c}$,$\frac{1}{d}$,$\frac{1}{e}$ are in A.P prove that a,c,e are in G.P.

Question

jiteshmeghwal9 · Answer

@ganeshie8

jiteshmeghwal9 · Answer

@priyar

anonymous · Answer

Maybe turn the G.Ps into equations? $$
\frac cb = \frac dc
$$

anonymous · Answer

$$
b-a = c-b
$$And $$
\frac 1d - \frac 1c = \frac 1e -\frac 1d
$$

jiteshmeghwal9 · Answer

@wio plez tell from which eqn i should start ?

jiteshmeghwal9 · Answer

i think $$\frac{c-d}{b-c}=\frac{c+d}{b+c}$$

parthkohli · Answer

Let the AP be $a, a+k, a+2k$. Now given the second statement$$d = \frac{(a+2k)^2}{a+k}$$Also$$\frac{1}e = \frac{2}d - \frac{1}c = \frac{2(a+k)}{(a+2k)^2} - \frac{1}{a+2k}=\frac{a}{(a+2k)^2}$$So$$a = a$$$$c = a+2k$$$$e = \frac{(a+2k)^2}a$$Do you see it now? :)

jiteshmeghwal9 · Answer

@ParthKohli how did u gt $$d=\frac{(a+2k)^2}{(a+k)}$$

parthkohli · Answer

$a_1, a_2, a_3$ in GP:$$a_2^2 = a_1 a_3$$or$$a_3 = \frac{a_2^2}{a_1}$$

jiteshmeghwal9 · Answer

Thanx @ParthKohli now i gt it :)